An electron acquires 13.0 × 10-16 j of kinetic energy when it is accelerated by an electric field from plate a to plate b. what is the potential difference between the plates?

solution:

the kinetic energy acquired by the electron when it is accelerated y an electric field is

k=5.25\times10^-16j

the kinetic energy acquired by the electron will ne proportional to the potential difference between the plates across which the electric field is applied.it is given by,

e\Delta v = k

here,\delta v is the potential difference between the plates, e is the charge on an electron and k is the kinetic energy.

reassange the above expression

\delta v = \frac{k}{e}

the potential difference between the plates will be  

\delta v = \frac{k}{e}

sunstitute 5.25 \times 10^-16j for k and 1.6 \times10^-19c for e is the above equation

\delta v =\frac{5.25\times10^-16j}{1.60\times10^-19c}

=3.28\times10^3v(\frac{1kv}{1000v})

=3.28kv

since the electrons will e accelerated towards the plate at higher potential.

hance p;ate b will be at higher potential

12000j is work done

where is the circuit diagram ?